The equation `|(x)/(x-1)|+|x|=(x^(2))/(|x-1|)`has

Options

A. exactly one solutions
B. exactly two solutions
C. at most two solutions
D. infinite number of solutions

Explanation

To solve the equation \( \left| \frac{x}{x-1} \right| + |x| = \frac{x^2}{|x-1|} \), we will analyze the equation step by step.

### Step 1: Analyze the absolute values
We need to consider the cases for the absolute values involved in the equation. The absolute value expressions depend on the sign of \( x \) and \( x-1 \).

1. **Case 1**: \( x > 1 \)
- Here, \( |x| = x \) and \( |x-1| = x-1 \).
- The equation becomes:
\[ \frac{x}{x-1} + x = \frac{x^2}{x-1} \]

2. **Case 2**: \( 0 \leq x \leq 1 \)
- Here, \( |x| = x \) and \( |x-1| = 1-x \).
- The equation becomes:
\[ \frac{x}{1-x} + x = \frac{x^2}{1-x} \]

3. **Case 3**: \( x < 0 \)
- Here, \( |x| = -x \) and \( |x-1| = 1-x \).
- The equation becomes:
\[ -\frac{x}{1-x} - x = \frac{x^2}{1-x} \]

### Step 2: Solve Case 1: \( x > 1 \)
For \( x > 1 \):
\[ \frac{x}{x-1} + x = \frac{x^2}{x-1} \]
Multiplying through by \( x-1 \) (which is positive):
\[ x + x(x-1) = x^2 \]
Simplifying:
\[ x + x^2 - x = x^2 \]
This simplifies to \( 0 = 0 \), which is true for all \( x > 1 \).

### Step 3: Solve Case 2: \( 0 \leq x \leq 1 \)
For \( 0 \leq x \leq 1 \):
\[ \frac{x}{1-x} + x = \frac{x^2}{1-x} \]
Multiplying through by \( 1-x \) (which is non-negative):
\[ x + x(1-x) = x^2 \]
Simplifying:
\[ x + x - x^2 = x^2 \]
This simplifies to:
\[ 2x - 2x^2 = 0 \]
Factoring gives:
\[ 2x(1-x) = 0 \]
Thus, \( x = 0 \) or \( x = 1 \).

### Step 4: Solve Case 3: \( x < 0 \)
For \( x < 0 \):
\[ -\frac{x}{1-x} - x = \frac{x^2}{1-x} \]
Multiplying through by \( 1-x \) (which is positive):
\[ -x(1-x) - x(1-x) = x^2 \]
This simplifies to:
\[ -x + x^2 - x = x^2 \]
Thus:
\[ -2x = 0 \implies x = 0 \]
However, \( x = 0 \) does not satisfy \( x < 0 \).

### Summary of Solutions
- From Case 1, we have infinitely many solutions for \( x > 1 \).
- From Case 2, we have two solutions: \( x = 0 \) and \( x = 1 \).
- Case 3 yields no valid solutions.

### Final Answer
The equation has infinitely many solutions for \( x > 1 \) and two specific solutions at \( x = 0 \) and \( x = 1 \).